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Suppose $(X,\mathcal{F},\mu,T)$ is an ergodic measure preserving dynamical system. Let $Y\subset X$ be such that $\mu(Y)>0$ and suppose there is an integrable function $R:Y\to \mathbb{N}$ such that $T^{R(y)}(y)\in Y$.

Then we can define a function $F:Y\to Y$ by $F=T^R$ and consider the induced system $(Y,\mathcal{F}\cap Y, \mu|_Y,F)$.

Can we say that $F$ is ergodic?

When $R(x)=\inf\{n\ge 1 : T^n(x)\in Y\}$, this induced system is very well studied and the answer to my question is yes (see for example http://www.weizmann.ac.il/math/sarigo/sites/math.sarigo/files/uploads/ergodicnotes.pdf, Theorem 1.7). However, the given proof does not extend to the general return times I am considering above.

Thanks :)

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In general $F$ is not ergodic. A very simple example can be constructed as follows: let $X=\mathbb{Z}_3=\{0,1,2\}$ and $\mu =1/3(\delta_0+\delta_1+\delta_2)$ and $T(x):=x+1$. This is an ergodic system. Let us define $Y:=\{0,1\}$ and $R\equiv 3$. Since $F=T^3=id$, it is not ergodic.

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    $\begingroup$ Is there any counter example with $T$ totally ergodic? $\endgroup$
    – KhashF
    Mar 23 '20 at 20:39
  • $\begingroup$ Isn't $T$ totally ergodic in Alejandro's counter example? $\endgroup$
    – Joël
    Mar 24 '20 at 0:45
  • $\begingroup$ @Joël, Doesn't totally ergodic mean that all iterates are ergodic? That certainly is not the case if $T^3$ is identity. $\endgroup$
    – KhashF
    Mar 24 '20 at 1:45
  • $\begingroup$ Totally ergodic means that there exists only one $T$-invariant measure (it is a property of the space, the $\sigma$-algebra, and $T$, but not of the measure $\mu$) $\endgroup$
    – Joël
    Mar 24 '20 at 2:06
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    $\begingroup$ @Joël I think that's unique ergodicity: en.wikipedia.org/wiki/Ergodicity#Unique_ergodicity $\endgroup$
    – KhashF
    Mar 24 '20 at 2:22
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I don't think that this system is even automatically measure-preserving (unlike the traditional induced map, which as you note is measure-preserving and ergodic).

Just take something silly like $(X, T)$ an irrational circle rotation with Haar measure, $Y$ the left half $[0, 1/2)$, and $R$ the first return time of a point in $Y$ to $[0, 1/4)$ (this always exists by minimality).

Then $T^R(y) \in [0, 1/4))$ for all $y \in [0, 1/2)$, so, for instance, $\mu|_Y([0, 1/4)) = 1/2$, but $\mu|_Y\big((T^R)^{-1}([0, 1/4))\big) = \mu|_Y(Y) = 1$. (I am here normalizing $\mu|_Y$ to make it a probability measure.)

I know my example is silly in that it's not even surjective on $Y$, but this isn't the problem; you could make a slightly trickier example where $R$ is the first return time to $[0, 1/4)$ for $y \in [0, 3/8)$ and the first return time to $[1/4, 1/2)$ for $y \in [3/8, 1/2)$, and then $T^R$ is surjective on $Y$, but still not measure-preserving.

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